Hi guys. First time posting, and I really need some help on some physics questions. We recieved some review of about 40 questions for our exam(tomorrow!!), and these 3 have stumped me. 1) If there are two points on a conductor each with the same potential...what is the signifigance of these two points? And how much work would be required to move a charge between these two points? 2) A roller coaster reaches the top of a grade at 10 m/s. It then goes downhill and uphill once more. If the second peak is 15m lower than the first, how fast will the roller cosaster be moving at that peak(neglecting friction)? 3) If an electrion moving downards past a magnet's south pole, what will be the electron beam's path as it mvoes by the south pole? Help would be appericatd. Thanks.
look up the definition of potential, and how its related to work. This is definitly a conservation of energy problem,, show what you've done so far, no one will do the work for you. Which way does the magnetic field point near the south pole. What will be the force on the electron as a result of this magnetic field ( The equation involves cross products )
1) Potential is the amount of potential energy per coloumb. 1 joule of energy is required to move one coulomb of charge through an electric field. Would that be the answer they are looking for? Also, what do they mean by "discuss the signifigance of these two points"? Would these two points be the poles or something? 2) I think I got the answer to this one by using e=mgh and e=1/2mv^2. So thanks! 3) Im not sure about all this. All I have is a diagram that shows an electron heading downwards, and the question "Describe the Path of the electron beam as it passed by the south pole". Also, I got another question that is me. 4)A box with an inital speed of 2m/s slides to a rest on a horizontal floor in 3 metres. What is the coefficient of kinetic friction? What I have done: found time. d=[(vi + v2)/2](t) 3=(2/2)(t) t=3s Found acceleration to be 2/3 m/s^2 But i have no idea where to go now. I dont know how to find friction coefficients without being given any forces!
Also, a 5th question. I thought I could figure out this question, but I cant! A stone is dropped into a well. The sound of the splash is hear 3s after the stone is dropped. What is the depth of the well. For some reason, the q doesnt give me a temperature to figure out the speed of sound.. so i just assumed it was 0 degrees celsius. What I did was: I used sound equations and kinematic equations. Vs=332 m/s I knew that the time it took for the stone to hit the water and for a sound to come back would be 3 seconds. So t1(time for it to hit water) + t2(time it took for sound to come up)=3s t1+t2=3s d=v1*t1 + 1/2*9.81*t1^2 d=4.905*t^2 (d/4.905)^1/2=t1 d=Vs*t2 d/332=t2 so...t1+t2=3=(d/332) + [(d/4.905)^1/2] I do all that, and eventually come to a quadratic equation. And then it just gets really complex, and I get a completley wrong answer(like 870000 m). What am I doing wrong?
What I was hoping you would recognize (and maybe your book doesn't mention this ) is that Work is equal to the change in potential energy, that is [tex] \vert \abs{W} \vert = \vert \Delta U \vert \ \ \ \ \ \ \ \ (1)[/tex] as you mentioned potential is the potential energy of a unit charge, therefore [tex] V = \frac{U}{q} [/tex] [tex]\Rightarrow \Delta V = \frac{ \Delta U }{q} [/tex] if we devide (1) by q, then [tex] \vert \abs{ \frac{W}{q} } \vert = \vert \frac{\Delta U}{q} \vert [/tex] [tex]\Rightarrow \vert \frac{W}{q} \vert = \vert \Delta V \vert [/tex] Thus [itex] \Delta V [/itex] can be thought of the work done on a unit charge by an electric field. I think figuring out the second part of the problem first will help you with this question. By convention, the magnetic field [itex] \vec{B} [/itex] points toward the south end of a magnet. In order for the electron to change its path, it needs to have an acceleration, so by newton's second law it needs to have a force acting on it. What is the equation of the force acting on a point charge by a magnetic field?, Its [tex] \vec{F} = q \vec{v} \ X \ \vec{B} [/tex] So if you can find the direction of the force, you know the direction of acceleration, and thus the change in the direction of the e's velocity. (Do you know how to find the direction of a cross product? ) Be careful about signs, the equation for acceleration is [tex] a = \frac{v_f - v_i}{t} [/tex] [tex]\Rightarrow a = \frac{ 0 - 2 }{3}[/tex] [tex]\Rightarrow a = - \frac{2}{3} \frac{m}{s^2} [/tex] Set up a Free-Body-Diagram for the box. Then set up Newtons second law equations for the vertical and horizontal components of force. You have the horizontal component of acceleration so see if you can solve what the force of fricion is... then you can use the uquation for friction to find the coefficient.
I had trouble understanding questions 2 and 3, but thankfully nothing on them was on the exam!! Right now I have to study for my math exam, but I'll be sure to read over your post again once im done that(for next year)! Thanks man!